Computing $\lim\limits_{n\to+\infty}\sum_{p=n+1}^{kn} \frac1 p$ for $k > 1$
Prove that
$$ \forall x>0,\quad \frac{1}{1+x}< \ln(1+x)-\ln(x)<\frac1 x.$$
Deduce, for $k\in \mathbb{N}\setminus\{1\}$,
$$\lim\limits_{n\to+\infty}\sum_{p=n+1}^{kn}\frac{1}{p}.$$
Show Answer:
We apply the mean value theorem to the function $t\mapsto \ln t$ on the interval $[x,x+1]$. It exists $c\in (x,x+1)$ such that
$$ \ln(1+x)-\ln x= \frac1 c.$$
Since $x < c < x+1$, we have
$$\frac1{x+1}<\frac1 c<\frac1 x,$$
and we get the desired result.
The bounds
$$\lim\limits_{n\to+\infty}\sum_{p=n+1}^{kn} \ln(p+1)-\ln p \le \lim\limits_{n\to+\infty}\sum_{p=n+1}^{kn}\frac1 p\le \lim\limits_{n\to+\infty}\sum_{p=n+1}^{kn} \ln p-\ln(p-1)$$
gives us by telescoping
$$\ln \frac{kn+1}{n+1}\le \lim\limits_{n\to+\infty}\sum_{p=n+1}^{kn}\frac1 p\le \ln k.$$
Finally, by the squeeze theorem
$$\lim\limits_{n\to+\infty}\sum_{p=n+1}^{kn} \frac1 p=\ln k.$$